Question

In the card game Poker each player is dealt a hand of 5 cards from a deck of 52, consisting
of 13 denominations (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) in each of 4 suits (♥, ♠, ♦, ♣). Suppose
each Poker hand is equally likely to occur. What is the probability that a Poker hand contains exactly three cards of denomination Q (as in the hand A♠, Q♦, Q♠, Q♥, 10♦, for example)?

Answer 1

Write your answer in terms of combinations.

Answer 2

@SithsAndGiggles Here is another one for you when you get a chance. :) Thanks!

Answer 3

@raffle_snaffle

Answer 4

\[\large P(3\ queens)=\frac{C(4, 3) \times C(48, 2)}{C(52, 5)}\]

Answer 5

I get
(1C1 4C3 12C2 4C1 4C1)/52C5

Answer 6

Hm, that seems to be the same, just written differently,, right @kropot72 ?

Answer 7

No, actually it looks a little different...

Answer 8

Sorry, I can't follow your solution. My solution uses the hypergeometric distribution, where there are 4 queens in the pack and a 5 card hand is dealt. I take it that the solution does not require that probability of exactly the example hand being dealt.

Answer 9

....the probability...*

Answer 10

Right. I think I see. I am trying to keep the hand from being a full house, but I don't think I have to. As it doesn't require only a three of a kind.

Answer 11

Just 3 queens.

Answer 12

Yes, there is no requirement on cards other than queens in the hand. Also, any suit can be excluded from the queens in the hand.

Answer 13

Right. Thank you so much!

Answer 14

You're welcome :)